Question: Solve for $y$, $ \dfrac{3}{16y + 8} = -\dfrac{y + 8}{4y + 2} + \dfrac{5}{16y + 8} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $16y + 8$ $4y + 2$ and $16y + 8$ The common denominator is $16y + 8$ The denominator of the first term is already $16y + 8$ , so we don't need to change it. To get $16y + 8$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{y + 8}{4y + 2} \times \dfrac{4}{4} = -\dfrac{4y + 32}{16y + 8} $ The denominator of the third term is already $16y + 8$ , so we don't need to change it. This give us: $ \dfrac{3}{16y + 8} = -\dfrac{4y + 32}{16y + 8} + \dfrac{5}{16y + 8} $ If we multiply both sides of the equation by $16y + 8$ , we get: $ 3 = -4y - 32 + 5$ $ 3 = -4y - 27$ $ 30 = -4y $ $ y = -\dfrac{15}{2}$